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Consider $$S_n = \sum_{i=0}^n a_i$$ and its Cesàro sums, defined as $$ C = \lim_{n \to \infty} \frac1n\sum_{k=0}^n S_k$$ Is it always true that $$ C = \lim_{n \to \infty} \frac1{L(n)}\sum_{k= n - L(n)}^n S_k$$ where $L(n)$ is any strictly increasing function such that $ 2 < L(n) < \ln(n)$ for every $n$?

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    @User1952009 i think i wrote what i meant ? – 2017-02-06
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    Yes sorry , i edited – 2017-02-06
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    tell it to ClementC. Also do you have a precise $a_n$ in mind, or do you assume something about it an $S_n$, for example $O(1)$ or $O(n)$ ? – 2017-02-06
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    No specific a_n no. I wonder if it ALWAYS holds ; for all a_n. – 2017-02-06
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    Did you try things like $a_n = cos(\ln n) $ and $L(n) = 2$ ? – 2017-02-06
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    This cannot hold. Try $a_n=1$ if $n=3^k-k$ and $a_n=-1$ if $a_n=3^k$, for some $k\geqslant1$, thus the first terms of the sequence $(a_n)_{n\geqslant0}$ are $0|0|1|-1|0|0|0|1|0|-1|0$ and $S_n=1$ if $3^k-k\leqslant n<3^k$ for some $k\geqslant1$ while $S_n=0$ for every other $n$. In particular, $(S_n)$ is Cesaro summable with $C=0$ but, for $$L(n)=\lfloor\log_3(n)\rfloor$$ the ratios $$\frac1{L(n)}\sum_{k=n-L(n)}^nS_k$$ fluctuate between $0$ and $1$. – 2017-02-06
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    @Did: I included your comment (with attribution) in my answer -- is that alright? – 2017-02-06
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    @ClementC. Sure. (Maybe concentrate your post on its most relevant parts and eliminate everything else?) – 2017-02-06
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    @Did Thanks -- done. – 2017-02-06

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tl;dr: no.

Define for convenience $L(n)$; by assumption, we have $L(n)=o(n)$ as $n\to\infty$.

  • Note that if $L(n)$ is bounded, then this is clearly false as the few missing constant terms do not really matter: you cannot hope that $\sum_{k=0}^n S_k$ be bounded in general (that is, there are Cesàro summable sums which are not convergent in the usual summation sense). For instance, take $a_n=(-1)^n$.

  • Assuming now $L(n)\xrightarrow[n\to\infty]{} \infty$, this looks like you are wondering about some specific converse of the Stolz–Cesàro theorem.

However, as Did's comment above shows, this is still false even with this assumption: I reproduced this comment below:

This cannot hold. Try $a_n=1$ if $n=3^k-k$ and $a_n=-1$ if $a_n=3^k$, for some $k\geqslant1$, thus the first terms of the sequence $(a_n)_{n\geqslant0}$ are $0|0|1|-1|0|0|0|1|0|-1|0$ and $S_n=1$ if $3^k-k\leqslant n<3^k$ for some $k\geqslant1$ while $S_n=0$ for every other $n$. In particular, $(S_n)$ is Cesàro-summable with $C=0$ but, for $$L(n)=\lfloor\log_3(n)\rfloor$$ the ratios $$\frac1{L(n)}\sum_{k=n-L(n)}^nS_k$$ fluctuate between 0 and 1.

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    I edited the question , sorry i made a mistake – 2017-02-06
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    I see. Let me edit my answer, then. – 2017-02-06
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    @mick I have to go, and my new "answer" is incomplete. Maybe someone will drop by and finish it. – 2017-02-06
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    Tl;dr : no , what is that ?? – 2017-02-06
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    "Too long; didn't read: no" ("if you don't have time to read the proof, here is the answer: no"). – 2017-02-06
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    Is the answer finished now ?? – 2017-02-07
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    Yes, it is. ${}$ – 2017-02-07